Chemistry 101
In this experiment, a sample of rice vinegar was analyzed using titration with a standard
NaOH solution. We find the molarity of NaOH through using an oxalic acid to balance it and through stoichiometry. Through that we can figure out the molarity of an acid. In each bottle of a commercial vinegar, stated the percent of acidity it has per bottle. But we all know that this percentage is just an approximation. Through titration we can figure out how much is the real percentage of acid in a bottle.
We did three trial for this experiment. We diluted the NaOH in experiment 12 and because of that we also have to dilute the acetic acid to save the amount of NaOH used in every titration. The solution NaOH was diluted to 8.3mL of 6M of NaOH to get 0.1M NaOH and placed in a flask with the oxalic acid and phenolphthalein until it got balanced.The calculated molar concentration of acetic acid was 0.06525M,average of three trials, which had a 3.9% acetic acid. Which this gave a 2.5 % difference compared to the 4.0% concentration reported on the bottle of the rice vinegar. There are some possible errors while doing the titration that led me to get this percentage, one possible source of error would be transferring of solutions, this will cause for me to loose some of the solution, it can also be an error in the records, my initial reading says 0.00mL but perhaps its 0.1mL, maybe 0.2mL or more. Lastly, probably the mistake in trial 1, I over shoot it and became dark pink, resulting in an incorrect calculated molarity for acetic acid.